1. Let S be between R and T. Use the Segment Addition Postulate to solve for m.
RS = 2m + 2
ST = 3m + 3
RT = 15
A. m = 3
B. m = 2
C. m = 9
D. m = 4
D.4
To solve for m, we can use the Segment Addition Postulate, which states that the sum of the lengths of two adjacent segments is equal to the length of the whole segment.
Given that RS = 2m + 2, ST = 3m + 3, and RT = 15, we can set up the equation:
RS + ST = RT
(2m + 2) + (3m + 3) = 15
Simplifying the equation, we have:
2m + 2 + 3m + 3 = 15
Combining like terms, we get:
5m + 5 = 15
Subtracting 5 from both sides, we have:
5m = 10
Dividing both sides by 5, we get:
m = 2
Therefore, the correct answer is B. m = 2.
To solve for the value of m, we need to use the Segment Addition Postulate, which states that if S is between R and T, then RS + ST = RT.
Given:
RS = 2m + 2
ST = 3m + 3
RT = 15
Using the Segment Addition Postulate, we can set up the equation:
(2m + 2) + (3m + 3) = 15
Combine like terms:
5m + 5 = 15
Subtract 5 from both sides of the equation:
5m = 10
Divide both sides of the equation by 5:
m = 2
Therefore, the answer is B. m = 2.
RT = RS + ST = 15.
(2m+2) + (3m+3) = 15.
m = ?.